1. 1.4.2 [5] <1.4> What is the global CPI for each implementation?
Consider two different implementations of the same instruction set architecture. There are four classes of instructions, A, B, C, and D. The clock rate and CPI of each implementation are given in the following table.
1.4.1 [10] <1.4> Given a program with 106 instructions divided into classes as follows: 10% class A, 20% class B, 50% class C, and 20% class D, which implementation is faster?
1.4.2 [5] <1.4> What is the global CPI for each implementation?
1.4.3 [5] <1.4> Find the clock cycles required in both cases.
Processor
Clock Rate
CPI (class A)
CPI (class B)
CPI (class C)
CPI (class D) P1
2.5 GHz 1 2 3 P2
3 GHz

2. Class A (10% of 106 instr.): 105 instr.
Solution 1.4.1
Class A (10% of 106 instr.): 105 instr.
Class B (20% of 106 instr.): 2 × 105 instr.
Class C(50% of 106 instr.): 5 × 105 instr.
Class D(20% of 106 instr.): 2 × 105 instr.
CPU-Time = AƩD(ICi x CPIi) x Clock Cycle Time = (IC x CPI)/Clock Rate
Used for:
1. CPU-Time(P1) =
( × 105 × × 105 × × 105 × 3)/(2.5 × 109) = × 10−4 s
CPU-Time(P2) =
(105 × × 105 × × 105 × × 105 × 2)/(3 × 109) =
6.66 × 10−4 s
Return

3. Global CPI = (CPU-Time x Clock Rate)/IC Therefore:
Solution 1.4.2
Global CPI = (CPU-Time x Clock Rate)/IC
Therefore:
CPI (P1) = 10.4 × 10−4 × 2.5 × 109/106 = 2.6
CPI (P2) = 6.66 × 10−4 × 3 × 109/106 = 2.0
Return

4. Class A (10% of 106 instr.): 105 instr.
Solution 1.4.3
Class A (10% of 106 instr.): 105 instr.
Class B (20% of 106 instr.): 2 × 105 instr.
Class C(50% of 106 instr.): 5 × 105 instr.
Class D(20% of 106 instr.): 2 × 105 instr.
CPU-Time = AƩD(ICi x CPIi) x Clock Cycle Time
 Clock Cycles = (AƩD(ICi x CPIi)
Therefore:
clock cycles (P1) = 105 × × 105 × × 105 × × 105 × 3 = × 105
clock cycles (P2) = 105 × × 105 × × 105 × × 105 × 2 = 20 × 105
Return